Question

3a. Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include...

3a. Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all charges, stoichiometric coefficients, and phase subscripts. (4 pts)

c. Determine [Ag+1] and [CrO4−2] fo a saturated solution of Ag2CrO4 if KSP = 1.1 × 10−12. Also, determine the molar solubility. (8 pts)

d. Use the KSP above to determine [CrO4−2] for a solution that is saturated with Ag2CrO4, and also has a common ion effect where [Ag+1] = 0.495 M. (4 pts)

Homework Answers

Answer #1

(a)

Solubility equation is,

Ag2CrO4 (s) 2 Ag^+ (aq) + CrO4^2- (aq)

Solubility product expression can be written as,

Ksp = [Ag^+]^2[CrO4^2-]

(b)

If solubility of Ag2CrO4 is S M, then [Ag^+] = 2 S M and [CrO4^2-] = S M

Ksp = S^2 x S

Ksp = S^3

S = (Ksp)^(1/3)

S = (1.1 x 10^-12)^(1/3)

S = Solubility of Ag2CrO4 = 1.03 x 10^-4 M

From the solubility equation,

[Ag^+] = 2 x [Ag2CrO4] = 2 x 1.03 x 10^-4 = 2.06 x 10^-4 M

[CrO4^2-] = [Ag2CrO4] = 1.03 x 10^-4 M

(c)

Solubility product expression can be written as,

Ksp = [Ag^+]^2[CrO4^2-]

1.1 x 10^-12 = (0.495)^2[CrO4^2-]

1.1 x 10^-12 = 0.245 [CrO4^2-]

[CrO4^2-] = 1.1 x 10^-12 / 0.245

[CrO4^2-] = 4.08 x 10^-12 M

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